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19

The proper technical term for the quantity is velocity, and it is a vector quantity. Speed is a colloquial name. If physicists use it, they only use it for the magnitude of velocity, but most of the time there is absolutely no reason to use it because the quantity is a vector one and acts as a vector one in all equations. Now square of a vector quantity is ...


11

Yes they are dimensionless numbers, which does not mean that they are constants. $C_L$ $C_D$ are variables. Dimensionless meaning: no physical unit. $$L = C_L \cdot \frac{1}{2} \rho V^2 \cdot A$$ with metric units: L [N] = [kg*m/sec$^2$] $\rho$ [kg/m$^3$] V [m/sec] A [m$^2$] Dimension of $\rho V^2 \cdot A$ = $\frac{kg}{m^3} \cdot \frac{m^2}{s^2} \cdot m^2 =...


8

Stalling means that flow separation increases so lift does no longer grow with angle of attack. This flow separation happens mostly over the rear, inner part of the wing, where the local surface inclination has a backward component. Now compare the pressure over an airfoil for fully attached and partially separated flow, just as it looks near or at stall. ...


8

While it’s not a true air to air fighter, following are some comparison numbers for the EA-6B Prowler that might help provide a useful ratio for a tactical "fighter type" aircraft. (Information comes from the NATOPS pocket checklist, under the constant altitude “bingo” fuel divert tables.) Theses figures are for a flaps-up aircraft, fully loaded ...


7

Amazing! British pioneer JW Dunne proposed this as a propulsion system around 1901 but Sir Hiram Maxim told him it would not work. (ref. unpublished documents in the Science Museum Archive's Dunne collection). Nine years later Henri Coanda built a ducted "jet" plane on broadly similar principles, which failed miserably (The Wikipedia article is not ...


7

In a way, yes. By using smooth metal surfaces with countersunk rivets, modern aircraft use a surface with low friction already. In the past, wooden structures were varnished to give them a smoother surface. In the second World War, German crews would polish the normally matte paint of some of their airplanes to eke out the maximum in top speed. There have ...


6

Let's imagine 2 theoretical wings, both of which have the same area, but differ in aspect ratio. Then the wing with the higher aspect ratio also has more span. This is what counts. If induced drag depends on the downwash angle, why would a longer wingspan reduce the angle? Because the wider wing will affect more air. Think of the air affected by the wing ...


5

Because velocity (being a vector) is a little more precise than pure (scalar) speed - so the angle between the object (aircraft, airfoil, ...) and the incoming air is also taken into consideration.


5

Thank you for sharing your calculation! Now I can see how you arrived at the results. Never mind the length: All that detail is essential for improving your results. Regarding the Breguet equation I prefer to use the one here and here. This, using your data for the A319, gives $$\frac{L}{D} = \frac{g\cdot b_f\cdot R}{v\cdot ln\left(\frac{m_2}{m_1}\right)} = ...


5

$C_D$ is a form factor variable, constructed to be as independent as possible of all the factors that contribute to aerodynamic forces. Two dimensionless flow parameters do influence outcomes of aerodynamic measurements: The Mach number, which quantifies compressibility of air. The Reynolds number, which accounts for viscosity/inertia effects in the ...


4

My source is STUDIES OF HIGH LIFT/DRAG RATIO HYPERSONIC CONFIGURATIONS by John V. Becker, to be found online here. It is from 1964, so more than 50 years old, but since a lot of research had been performed already before that date, it might still be relevant. In short: It depends. On thickness ratio and Reynolds number, for example, as can be seen in this ...


4

I did some flight testing last night in a C182, and I saw a speed increase of 3-4 knots when I closed the cowl flaps at cruise. While I won’t pretend my methodology was rigorously scientific, the results were consistent enough across different flight profiles that I’m confident in saying FlightGear’s 20 knots is grossly incorrect. The C210 cruises quite a ...


4

Simply using the generic drag equation will get you within the ballpark required for FlightGear. $$D = C_D \cdot \frac{1}{2} \rho V^2 \cdot A$$ with $C_D$ the flat plate drag coefficient and A being a reference area of your cowl flaps. The data for the correct Reynolds number is best used, with $$Re = \frac{\rho V \bar{c}}{\mu}$$ with $\bar{c}$ = mean ...


3

Yes, you chose the wrong airfoil. While the NACA 6-digit series was among the first set of airfoils computed from a design pressure distribution, they will suffer from shocks when operated above their critical Mach number just as any other airfoil. Comparison of drag rise over Mach for 6-series and early supercritical airfoils from NASA Technical Paper 2969....


3

The downwash is about as high as wide, and angled only by a few degrees (more at slow speed). So the longitudinal distance needs to be many times longer than the wing span to make the wings independent. And it would still be less efficient than increasing the span: If you double the span, you will decrease induced drag four times. If you half the lift, you ...


3

You seem to be confusing force (thrust or drag) with energy (under which concept work and power comes). Don't. You'll be spared much grief. Work done is force times distance. $$ W = Fs $$ Power is work done divided by time $$ P = W/t $$ When you double velocity, force quadruples. As does work. $$ W_\text{new} = 4Fs = 4W $$ Time halves as well, with the ...


3

Could drag in a turn be twice the drag in straight and level flight? Yes, but this will be a rater steep turn. … in a coordinated turn with the same angle of attack at the same airspeed … No, this is impossible. Either angle of attack or speed have to be higher so the higher lift required for turning can be generated. … angle of attack is the same, the ...


3

Induced drag $D_i$ is proportional to the square of lift $L$: $$D_i = \frac{2\cdot L^2}{\rho\cdot v^2\cdot\pi\cdot b^2\cdot\epsilon}$$ where $b$ is wingspan, $v$ is flight speed and $\rho$ is air density. $\epsilon$ is an efficiency factor which tends to be between 0.8 and 1. Compared to straight and level flight, induced drag will quadruple when turning at ...


3

The technical report NACA-TR-824 has been digitized. The digitized version includes Cl and Cd for 118 airfoils at 3 Reynolds numbers. To get it, create an account on x-plane.org. (It's been online sporadically elsewhere over the decades, e.g. here. People associated with the digitizing include Gregory Peter, Gregory Siemens, and James Sonnenmeier.) ...


3

It’s just because it is derived from a larger more complete theory of aerodynamic forces where it’s necessary to make the distinction. In this equation the coefficient is simplified but a more complete calculation of the coefficient uses the directional components of velocity. https://www.grc.nasa.gov/www/k-12/airplane/eulereqs.html


3

Concorde afterburners were reducing overall fuel consumption. To understand why there is a special drag penalty around Mach 1, please read this answer. Now to your graphs in the question body. The first one with the steep drag peak at Mach 1 is for a straight wing which never was designed to fly trans- or supersonically. You do get such results, but only if ...


3

For this answer I have to speculate since I do not have the wind tunnel data which would be needed for a more substantial answer. Why should adding the spoiler reduce drag at lower speed? The most likely reason is a changed flow separation over the car's rear area. I would expect that flow separates already at some point along the rear window if no spoiler ...


2

Generally, you are right. Reducing wing area is reducing overall drag. Within limits. Induced drag depends on speed and span loading. If the reduced wetted surface allows you to fly faster, reduced drag will be lower, leaving more of the power budget to overcome viscous drag. However, if wing span is reduced, induced drag will be higher at the same speed, so ...


2

Lowering flaps and gear will add significant drag, which causes the plane to decelerate. Your body feels that, but without visual reference to the ground, your brain has no way to know what speed it was traveling before or after that deceleration. This is an example of “spatial disorientation”, where the brain gets confusing inputs and comes to the wrong ...


2

It's not so simple. For swept wings it is essential whether the leading edge is sub- or supersonic. What does that mean? If the leading edge is within the Mach cone emanating from the tip of the wing, it is called subsonic. The approaching air will "sense" what is coming and behave much like in subsonic flow. This avoids wave drag and keeps nose ...


2

No, especially not for the conditions you stated. Even if we consider that "same exact ariplane" will not melt away at Mach 2 or 5, different phenomena start to appear at such speeds. Let's restrict speeds to below about Mach 0.5 for now. At such speeds, compressibility of air does not affect characteristics significantly, and things become easier. ...


2

You could argue for one or for both, but ultimately, it doesn't matter - as long as you are consistent and use the same reference area throughout your calculation/analysis. If I personally were to choose, I would go for both, on the basis of total lifting area. EDIT for details ------ Well yet again, it doesn't matter. You will get some value for your $C_{...


2

Your lift will be prescribed and cannot be freely chosen. So what you need to do is to minimize the drag that goes with this given lift. In straight flight lift needs to equal weight. In turns, the centripetal force is added to this. At high speed a small angle of attack will produce that lift while at low speed the angle of attack needs to be high. In most ...


1

You would solve this problem by drawing the lift vector being tilted back at half the downwash angle, then apply some trigonometry to find the horizontal bit of the lift vector.


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